AMC 10 · 2014 · #25

Grade 8 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicrecursive-sequencesystems-of-equationssymmetry-argument convert-to-algebra ↑ Prerequisites: probability-basicsystems-of-equations

📏 Long solution 💡 4 insights

Problem

A frog starts on pad 1 of pads numbered 0 to 10. From pad N (with 0 ＜ N ＜ 10) it hops to N-1 with probability N/10 and to N+1 with probability 1-N/10. Pad 0 means eaten, pad 10 means escape. Find the probability the frog reaches pad 10 before pad 0.

Pick an answer.

(A)

$\frac{32}{79}$

(B)

$\frac{161}{384}$

(C)

$\frac{63}{146}$

(D)

$\frac{7}{16}$

(E)

$\frac{1}{2}$

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

There are eleven pads and one wandering frog, so instead of tracking every possible path we name one unknown per pad: s_N, the probability of escaping when currently on pad N (Tool #4, Introduce a Variable). One jump then links each s_N to its neighbors, turning the puzzle into a chain of equations. The board is mirror-symmetric about pad 5, so a change of focus (Tool #16) hands us s₅=1/2 for free. Solving the chain one pad at a time (Tool #7, Identify Subproblems) and then back-substituting (Tool #11, Work Backwards) pins down s₁.

1STEP 1

Name the escape chance per pad

Let s_N be the chance of escaping from pad N. Both ends are certain: s₀ = 0 and s₁₀ = 1, and the target is s₁.

s₀ = 0, s₁₀ = 1, find s₁

2STEP 2

Link each pad to its neighbors

Condition on one hop: s_N = N/10 s_N-1 + (10-N)/10 s_N+1, and clearing tenths gives 10 s_N = N s_N-1 + (10-N) s_N+1.

10 s_N = N s_N-1 + (10-N) s_N+1 (0 ＜ N ＜ 10)

3STEP 3

Use symmetry at the middle pad

Pad 5 is dead center: down and up are both 5/10, and N ↦ 10-N mirrors the whole board, so s₅ = 1/2.

s₅ = 1/2

4STEP 4

Climb from the bottom

With s₀ = 0, N=1 gives 10 s₁ = 9 s₂, so s₁ = 9/10 s₂; feeding that into N=2 leaves s₃ = 41/40 s₂.

s₁ = 9/10s₂, s₃ = 41/40s₂

5STEP 5

Reach the known middle value

N=3 yields s₄ = 29/28 s₂; putting s₃, s₄ and s₅ = 1/2 into 10 s₄ = 4 s₃ + 6 s₅ leaves 219/35 s₂ = 3, so s₂ = 35/73.

219/35 s₂ = 3 → s₂ = 35/73

6STEP 6

Back-substitute for pad 1

Back-substitute: s₁ = 9/10 · 35/73 = 63/146, the escape chance from pad 1 — choice (C).

s₁ = 9/10·35/73 = 63/146 → (C)

Answer

63/146

63/146≈ 0.431 is exactly choice (C), and it is just under 1/2. That fits the story: the frog starts on pad 1, right next to the snake, so escape should be a bit less than even — yet the strong upward push there (up-probability 9/10) keeps it close. The relation 10 s₁ = 9 s₂ checks out: 9·35/73=315/73 equals 10·63/146=630/146=315/73. The symmetric value s₅=1/2 also lands between the endpoints as it must.

💡Key takeaway

Give every pad its own escape chance, use the fair 1/2 at the middle pad, then walk the equations up from the bottom to pin down pad 1.

Name the escape chance per pad
Link each pad to its neighbors
Use symmetry at the middle pad
Climb from the bottom
Reach the known middle value
Back-substitute for pad 1